SOLUTION: 1.FIND how many 5 digits odd number greater than 40000 that can be obtained using the digit 1,3,5,6,7,8. a) repetition of digits not allowed b) repetition allowed 2.6 men and

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Question 1082187: 1.FIND how many 5 digits odd number greater than 40000 that can be obtained using the digit 1,3,5,6,7,8.
a) repetition of digits not allowed
b) repetition allowed
2.6 men and 3 women are standing in a supermarket queue.
Find possible arrangements if NO TWO women are standing next to each other.

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
1.FIND how many 5 digits odd number greater than 40000
that can be obtained using the digit 1,3,5,6,7,8.
a) repetition of digits not allowed
Since the 5-digit numbers must be odd, their last digit is 1, 3, 5 or 7

Case 1.  The last digit is 1 or 3
Then the first digit is 5, 6, 7 or 8 in order to be greater than 40000.

We can choose the last digit either of 2 ways, as 1 or 3.
We can then choose the first digit any of 4 ways. (as 5, 6, 7 or 8.)
We choose the second digit as any of the remaining 4 digits.
We choose the third digit as any of the remaining 3 digits.
We choose the fourth digit as any of the remaining 2 digits.
Answer for case 1:  2×4×4×3×2 = 192 ways

Case 2.  The last digit is 5 or 7
We can choose the last digit either of 2 ways, as 5 or 7.
We can then choose the first digit any of 3 ways. 
       (as 6, 8, or the unchosen one of 5,7.)
We choose the second digit as any of the remaining 4 digits.
We choose the third digit as any of the remaining 3 digits.
We choose the fourth digit as any of the remaining 2 digits.
Answer for case 2:  2×3×4×3×2 = 144
Total for both cases:  192+144 = 336 ways.
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b) repetition allowed
Since the 5-digit numbers must be odd, their last digit is 1, 3, 5 or 7
We can choose the last digit any of 4 ways, as 1, 3, 5 or 7.
We can then choose the first digit any of 4 ways. (as 5, 6, 7 or 8.)
We choose the second digit as any of the 6 digits.
We choose the third digit as any of the 6 digits.
We choose the fourth digit as any of the 6 digits.
4×4×6×6×6 = 3456 ways
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6 men and 3 women are standing in a supermarket queue.
Find possible arrangements if NO TWO women are standing
next to each other.

First we arrange the 6 men in a line in 6! = 720 ways.

Think of the row of men like this with spaces between them,
1 space immediately before the leftmost man and 1 space 
immediately after the rightmost man, like this:

_M_M_M_M_M_M_

The 7 "_"'s are potential places where the three women can
be inserted among the men.  That's 7P3 = 7×6×5 = 210 ways to 
insert the 3 women in three of those blanks so that no two 
women are next to each other.

Answer: 6!×(7P3) = 720×210 = 151200 ways.
 
Edwin